Calcium carbonate reacts with aqueous `HCl` to give `CaCl_(2)` and `CO_(2)` according to the reaction:
`CaCO_(3)(s)+2HCl(aq) rarr CaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
What mass of `CaCO_(3)` is required to react completely with `25 mL` of `0.75 M HCl`?

Explanation

To solve the problem of how much calcium carbonate (CaCO₃) is required to react completely with 25 mL of 0.75 M HCl, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ \text{CaCO}_3(s) + 2 \text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2O(l) \]

### Step 2: Determine the moles of HCl
We need to calculate the number of moles of HCl present in 25 mL of 0.75 M solution. The formula for calculating moles from molarity is:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]

Convert the volume from mL to L:
\[ 25 \text{ mL} = 25 \times 10^{-3} \text{ L} = 0.025 \text{ L} \]

Now, calculate the moles of HCl:
\[ \text{Moles of HCl} = 0.75 \, \text{mol/L} \times 0.025 \, \text{L} = 0.01875 \, \text{mol} \]

### Step 3: Use stoichiometry to find moles of CaCO₃
From the balanced equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, we can set up the following relationship:
\[ \text{Moles of CaCO}_3 = \frac{\text{Moles of HCl}}{2} \]

Substituting the moles of HCl:
\[ \text{Moles of CaCO}_3 = \frac{0.01875 \, \text{mol}}{2} = 0.009375 \, \text{mol} \]

### Step 4: Calculate the mass of CaCO₃ required
To find the mass of CaCO₃, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]

First, we need to calculate the molar mass of CaCO₃:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 3 O atoms)

So, the molar mass of CaCO₃ is:
\[ 40 + 12 + (16 \times 3) = 40 + 12 + 48 = 100 \, \text{g/mol} \]

Now, calculate the mass:
\[ \text{Mass of CaCO}_3 = 0.009375 \, \text{mol} \times 100 \, \text{g/mol} = 0.9375 \, \text{g} \]

### Conclusion
The mass of CaCO₃ required to react completely with 25 mL of 0.75 M HCl is **0.9375 grams**.

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